View Full Version : begginer confused with numpy

jeremytaylor

May 4th, 2007, 04:56 PM

Hi, I'm a newcomer to python and want to use the numpy package for my work. I think I must still be confused about the differences between the way mutable and immutable objects are passed into functions so maybe someone can help me understand.

As far as I know numpy arrays are mutable. Therefore why, in the following code, does the array a not change its value?

from numpy import *

def f(x) :

y = array([[1, 2],[3,4]])

x = y.copy()

return None

a = array([[1, 1],[1,1]])

print a

f(a)

print a

Any help would be much appreciated.

Jeremy

WW

May 4th, 2007, 05:33 PM

Your array a refers to a block of data that holds the elements of the array. After calling f, a still refers to the same block of data; f does not change that. What happens in f is that initially, x refers to the same block of data as a, but when you say "x = y.copy()", x now refers to a different block of data; you have not changed the original block of data. When f returns, a still refers to the original block of data.

If you want the contents of a to be changed by the function f, don't reassign x to a new array; instead, changed the contents of x. For example, try this:

from numpy import *

def f(x) :

x[0,0] = 99

x[1,1] = -1

return None

a = array([[1, 1],[1,1]])

print a

f(a)

print a

baltimark

May 4th, 2007, 05:49 PM

Does this code clear things up at all?

from numpy import *

def f(x) :

y = x.copy()

y[0][0] = 3

return None

def g(x):

y = x

y[0][0] = 3

return None

def h(x):

x[0][0] = 5

return None

a = array([[1, 1],[1,1]])

print "a = \n" ,a

f(a)

print "after f, a = \n", a

g(a)

print "after g, a = \n", a

h(a)

print "after h, a = \n", a

You're probably trying to investigate the difference between functions f and g.

So, in 'f', I create a separate copy of x. That SHOULDN'T change 'a' when I change the copy. And it doesn't.

But, in 'g', the references are assigned.

In your example, x and a are the same thing when you go into the function, but then you assign x to something new. It doesn't carry 'a' along. That reference is just hanging now, really. It doesn't move the things from y into x's location. It moves x to somewhere else.

This guy takes a stab at it. . .

http://www.goldb.org/goldblog/CommentView,guid,4eb92070-c279-44b3-ac2a-5d1c4f3e8115.aspx

On that page, it says this

Alex is right that trying to shoehorn Python into a "pass-by-reference" or "pass-by-value" paradigm is misleading and probably not very helpful. In Python every variable assignment (even an assignment of a small integer) is an assignment of a reference. Every function call involves passing the values of those references.

jeremytaylor

May 4th, 2007, 06:00 PM

Thank you so much both of you! Starting to get the hang of it (too many years programming fortran eek)

Jeremy

dwblas

May 5th, 2007, 12:23 AM

I think this is what you want. You have to return the updated array, otherwise it remains local to the function and is destroyed when the function exits.

from numpy import *

def f(x) :

y = array([[1, 2],[3,4]])

x = y.copy()

return x

a = array([[1, 1],[1,1]])

print a

a = f(a)

print a

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